Two Pendulums connected by a spring

[CAF123]

Homework Statement

Identical pendulums of length $L$ are suspended from the ceiling a distance $a$ apart. Each of the pendulums carries a particle of mass $m$. The two particles are connected horizontally by a spring of natural length $a$. Consider only motion in the plane formed by the pendulums and the spring and assume that the displacments of the pendulums from the vertical, $x_1, x_2$ are small. (i.e the angles of deflection are small)

1)Express the total energy as a sum of kinetic and potential terms using the variables $x_1, x_2$ and their derivatives (NB: Include the potential energy stored in a spring and gravity)

2)Bring the energy into the form $$E = \frac{1}{2}\underline{\dot{x}^T} M \underline{\dot{x}} + \frac{1}{2}\underline{\dot{x}^T} K \underline{\dot{x}},$$ M, K matrices.

The Attempt at a Solution

1) I dealt with each pendulum separately. They are joined only via the spring and so this will come across in the changing elastic potential energy stored in the spring:
Define coordinates where the pendulum is attached to the ceiling with positive x to left and positive y downwards. Let $L = l$Then for this pendulum:
$$V_1 = -mgy_1 = -mg(\sqrt{l^2 - x_1^2}),\,\,\,T_1 = \frac{m}{2}\left[\frac{\dot{x_1}^2 l^2}{l^2 - x_1^2}\right]$$

For the other pendulum, again put x +ve to right and y postive down and origin at ceiling: Then $$V_2 = =mg\sqrt{l^2 - x_2^2}\,\,T_2 = \frac{m}{2}\left[\frac{\dot{x_2^2} l^2}{l^2 - x_2^2}\right]$$

For the spring, I think the expression will change depending on whether the springs move in phase or not. If one is displaced to the right by x1 and the other to the left by x2 then $U_S = \frac{1}{2}k(x_1 - x_2 - a)^2$

I can then add all these together. I haven't used the fact that the displacements are small. I thought about neglecting sqaured terms etc.. but then all the kinetic terms would drop out. I am not sure about my expression for $U_S$ either since I can come up with another expression depending on the relative motion of the two pendula.

Many thanks.

 

[voko]

It seems that the x1, x2 notation is not completely consistent. In the pendula' potential energy terms, x1 = 0 and x2 = 0 seem to denote the equilibrium position. But then the length of the string is x1 - x2 + a, yielding the extension as simply x1 - x2, not as x1 - x2 - a used in the spring's potential energy.

That said, I am not even sure you can assume that the displacement is purely horizontal when deriving the spring's potential energy. You don't assume that for the pendula. Hmmm.

 

[CAF123]

If x1 = x2 = 0 then the spring potential energy term should vanish surely?

But I am always going to have some sort of extension from the natural length so if I denote the extension of the spring by Δx, then I have Δx + a (why is it +?) which gives a spring potential energy of $\frac{1}{2}k(\Delta x + a)^2$ If $\Delta x$ is zero, then I get a spring potential energy of $\frac{1}{2}k(a)^2$ which doesn't make sense I don't think. (since at equilibrium, the pendula are relaxed so there is no spring potential energy).

Yes, I must have forgot to consider the y motion of the spring. Can you help me first getting the x component?

 

[voko]

If x1 = x2 = 0 then the spring potential energy term should vanish surely?

It does not have to. E.g., the gravitational terms are not zero at equilibrium. Potential energy is defined up to an additive constant.

But I am always going to have some sort of extension from the natural length so if I denote the extension of the spring by Δx, then I have Δx + a (why is it +?) which gives a spring potential energy of $\frac{1}{2}k(\Delta x + a)^2$

Just to make sure we are on the same page. "Extension" means the increase in the length of the spring in comparison to its natural length. So if Δx is the extension, then Δx + a is the simply the length of the (extended) spring.

Now the potential energy (as does the force) depends solely on the extension, not on the length, so it should be $\frac{1}{2}k(\Delta x)^2$. In this particular case we do have zero for the minimal potential energy.

Yes, I must have forgot to consider the y motion of the spring. Can you help me first getting the x component?

The problem is there are no x and y components in potential energy. It is not a vector.

 

[CAF123]

Just to make sure we are on the same page. "Extension" means the increase in the length of the spring in comparison to its natural length. So if Δx is the extension, then Δx + a is the simply the length of the (extended) spring.

Now the potential energy (as does the force) depends solely on the extension, not on the length, so it should be $\frac{1}{2}k(\Delta x)^2$. In this particular case we do have zero for the minimal potential energy.

Ok, but when I was writing out the hookian force F = -kx, I always had something of the form F = -k(x-a) where x is the extension from natural length a. (e.g in my last problem about coupled masses with transverse oscillations). Why don't we have something of the form 1/2 k (x-a)^2 here?

The problem is there are no x and y components in potential energy. It is not a vector.

For the kinetic energy terms, I wrote $T = \frac{m}{2}\left[\dot{x}^2 + \dot{y}^2\right]$ When the spring is moved via the movement of the pendula , there will be some movement in x and in y direction. Picturing the situation, it seems rather difficult to come up with an expression for the potential energy of the spring.

EDIT: Or could it be as simple as $$U = \frac{1}{2}k(x_1-x_2)^2 + \frac{1}{2}k(y_1 - y_2)^2?$$

 

[voko]

Ok, but when I was writing out the hookian force F = -kx, I always had something of the form F = -k(x-a) where x is the extension from natural length a. (e.g in my last problem about coupled masses with transverse oscillations). Why don't we have something of the form 1/2 k (x-a)^2 here?

Because x1 and x2 are defined differently here. If you had at equilibrium x1 - x2 = a (or x2 - x1 = a) then (x1 - x2 - a) would denote the extension. But you have x1 = x2 = 0 at equilibrium, so (x1 - x2 - a) cannot possibly be the extension.

It could be explained this way. x1 and x2 are defined relatively to the corresponding equilibrium points. We could define absolute coordinates X1 = x1, X2 = x2 - a. Then X1 - X2 = a at equilibrium, thus the extension is (X1 - X2 - a).

For the kinetic energy terms, I wrote $T = \frac{m}{2}\left[\dot{x}^2 + \dot{y}^2\right]$ When the spring is moved via the movement of the pendula , there will be some movement in x and in y direction. Picturing the situation, it seems rather difficult to come up with an expression for the potential energy of the spring.

Using the absolute coordinates defined above, the length of the spring is $\sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2}$, and so its extension is $\sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2} - a$, and the potential energy is $\frac 1 2 k (\sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2} - a)^2$.

 

[CAF123]

Using the absolute coordinates defined above, the length of the spring is $\sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2}$, and so its extension is $\sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2} - a$, and the potential energy is $\frac 1 2 k (\sqrt {(X_1 - X_2)^2 + (y_1 - y_2)^2} - a)^2$.

Or equivalently, $U = \frac{1}{2}k (\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2})^2$?
The expression I had in my edit (I am not sure if you caught it initally) summed together the potential energy in the x direction and the potential energy in the y direction. Is that also a satisfactory equation? I believe the one derived above takes some extension in the x direction and some extension in the y direction and computes the resulting extension via Pythagoras. Whereas the one I derived initially considers the cases separetely and I then add them together.

EDIT: I only just reliased now that both are exactly the same.

 

[voko]

Or equivalently, $U = \frac{1}{2}k (\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2})^2$?

I do not think it is equivalent to that. The first one has a clear geometrical and physical meaning, this one is not obvious at all.

It is actually equivalent to the sum of the potential energies you had, which does not seem justified, either.

Anyway, I think this is becoming too complicated so you could probably just ignore the vertical displacement for the spring and use the potential energy term you derived previously.

 

[CAF123]

So given the below as my energy:
$$E = -mg \sqrt{l^2 - x_1^2} + \frac{m}{2}\left[\frac{\dot{x_1}^2l^2}{l^2 - x_1^2}\right] -mg\sqrt{l^2 - x_2^2} + \frac{m}{2}\left[\frac{\dot{x_2}^2l^2}{l^2 - x_2^2}\right] + \frac{1}{2}k(x_1 - x_2)^2 + \frac{1}{2}k(y_1 - y_2)^2$$

Before I try to put this into the matrix form, I need to somehow use the small angle approx/small displacement. At the moment I have squared terms in $x_1,x_2,y_1,y_2$ and I think to put in matrix form they should be linear.

 

[CAF123]

I do not think it is equivalent to that. The first one has a clear geometrical and physical meaning, this one is not obvious at all.

It is actually equivalent to the sum of the potential energies you had, which does not seem justified, either.

Anyway, I think this is becoming too complicated so you could probably just ignore the vertical displacement for the spring and use the potential energy term you derived previously.

I derived it by considering the separate extensions in x and y and then summed them together. E.g some x1 -x2 in x direction and y1 - y2 in y direction. Why is this not correct?

When you say ignore the vertical displacement, do you mean since we have small displacements the spring stays near enough horizontal and so the potential is only $1/2 k (x_1 - x_2)^2$

 

[voko]

I do not understand why you are required to have two matrices. The expressions around M and K are identical, so you could just use N = M + K. Are you sure this what the problem is about? Perhaps one of the terms should be quadratic in displacements, not in velocities?

 

[CAF123]

I do not understand why you are required to have two matrices. The expressions around M and K are identical, so you could just use N = M + K. Are you sure this what the problem is about? Perhaps one of the terms should be quadratic in displacements, not in velocities?

There is an error in the problem statement. How could I be so careless.
It should read:
Bring the energy into the matrix form: $$\frac{1}{2}\underline{\dot{x}}^T M \underline{\dot{x}} + \frac{1}{2} \underline{x}^T K \underline{x}$$

 

[voko]

I derived it by considering the separate extensions in x and y and then summed them together. E.g some x1 -x2 in x direction and y1 - y2 in y direction. Why is this not correct?

Because extensions do not follow the Pythagoras scheme.

Assume you have right triangle with sides a and b. Its hypotenuse is $\sqrt {a^2 + b^2}$. Now, let the sides have some extensions A and B. Then the hypotenuse is $\sqrt {(a + A)^2 + (B + b)^2}$, and the extension of the hypotenuse is $\sqrt {(a + A)^2 + (B + b)^2} - \sqrt {a^2 + b^2} \ne \sqrt {A^2 + B^2}$.

When you say ignore the vertical displacement, do you mean since we have small displacements the spring stays near enough horizontal and so the potential is only $1/2 k (x_1 - x_2)^2$

Yes, that is what I mean. I know I said otherwise earlier :)

 

[voko]

There is an error in the problem statement. How could I be so careless.
It should read:
Bring the energy into the matrix form: $$\frac{1}{2}\underline{\dot{x}}^T M \underline{\dot{x}} + \frac{1}{2} \underline{x}^T K \underline{x}$$

This is better. So you have to retain the terms quadratic in velocities and displacements.

 

[CAF123]

This is better. So you have to retain the terms quadratic in velocities and displacements.

But I also have squared roots etc.. so how I can separate say $x_1^2$ from $\sqrt{l^2 - x_1^2}?$

 

[voko]

Taylor series. Retain the terms up to the power of two.

 

[CAF123]

Taylor series. Retain the terms up to the power of two.

Yes, so my expression becomes after simplification: $$-2mgl + mgx_1^2/2l + \frac{m}{2}\left[\frac{\dot{x_1}^2l^2}{l^2 - x_1^2}\right]+ mgx_2^2/2l + \frac{m}{2}\left[\frac{\dot{x_2}^2l^2}{l^2 - x_2^2}\right] + 1/2 kx_1^2 - kx_1x_2 + 1/2 k x_2^2$$

Set that first constant term = 0.

 

[voko]

You need to take care of the denominator in the kinetic energy terms.

 

[CAF123]

You need to take care of the denominator in the kinetic energy terms.

So I need to Taylor expand $l^2(l^2 - x_1^2)^{-1}$. I get then $1 + \frac{x_1^2}{l^2}$.

When I multiply this back onto $\frac{m}{2} \dot{x_1}^2,$ should I neglect the term that contains the coupled term $\dot{x_1}^2 x_1^2$. (It will be sufficiently small anyway).

 

[voko]

Yes, I think this is what you are supposed to do.

 

[CAF123]

Yes, I think this is what you are supposed to do.

The $mg/2l x_1^2$ term is now posing a difficulty because we want m to be encapsulated into M ( in the first term) and $x_1$ in the other term.

Also I should neglect the kx1x2 term that I get when I expand the potential term.

 

[voko]

The $mg/2l x_1^2$ term is now posing a difficulty because we want m to be encapsulated into M ( in the first term) and $x_1$ in the other term.

I do not follow.

Also I should neglect the kx1x2 term that I get when I expand the potential term.

Why? You are allowed to have such terms.

 

[CAF123]

I do not follow

Sorry, to be more clear in the first term I have M, so I presume all terms containing m have to go in that matrix. The above term contains m but there is no where for x1 to multiply on(since the expressions around M are $\dot{x}$ and not $x$

Why? You are allowed to have such terms.

I haven't found a way to put that term into the matrix yet (since the term has to be multiplied by both x1 and x2)

 

[voko]

I think you are attaching too much significance to letters :) You have to put the potential energy terms somewhere, and they will contain both m and k.

As for the x1 and x2 product, write down the explicit form of x^TKx.

 

[CAF123]

I see. Ok I will try this.

 

[CAF123]

Writing it out in full and multiplying out the matrix I get that M is simply $mI_2$, where $I_2$ is the 2x2 identity. For K, I have the top left entry as $mg/l + k$, the diagonal entries $k$ (which makes sense - if these had been zero it would not have made sense) and the bottom right entry $mg/l + k$ too.
(I don't know matrix latex yet so hopefully this is satisfactory).

The next question was:
3)Differentiate the expression for E and identify the eqns of motion. Show they take the form $\underline{\ddot{x}} + A\underline{x} = 0.$Give A.

When they say eqns, I presume that means more than one. The only ones I see at the moment are :$$\underline{\ddot{x}}^T M + \underline{x}^TK = 0 \,\,\,\text{and}\,\,\,M\underline{\ddot{x}} + K \underline{x} = 0$$ the latter being the familiar harmonic motion Eqn of motion. Are these the two the question is talking about?

 

[voko]

The first matrix equation you got is wrong. How did you get it? The second one is correct. Because it is a matrix equation, it is a system of equations. All you need to do is convert it to the form required.

 

[CAF123]

The first matrix equation you got is wrong. How did you get it? The second one is correct. Because it is a matrix equation, it is a system of equations. All you need to do is convert it to the form required.

$$dE/dt = 0 \Rightarrow \frac{1}{2}\underline{\dot{x}}^T M \underline{\ddot{x}} + \frac{1}{2}\underline{\ddot{x}}^T M \underline{\dot{x}} + \frac{1}{2}\underline{x}^T K\underline{\dot{x}} + \frac{1}{2}\underline{\dot{x}}^T K\underline{x} = \frac{1}{2}\underline{\dot{x}}[ \underline{\ddot{x}}^T M + \underline{x}^TK] + \frac{1}{2}\underline{\dot{x}}^T[M \underline{\ddot{x}} + K \underline{x}] = 0$$

I then identified the two eqns in the brackets.

 

[voko]

Ah, sorry, somehow I could not see the first equation had TWO dots over x! I thought it was just one.

Can you see the first equation is identical with the second one?

 

[CAF123]

Ah, sorry, somehow I could not see the first equation had TWO dots over x! I thought it was just one.

Can you see the first equation is identical with the second one?

Yes, the first one has just x transposed, so when I multiply it all out, I believe I get the same. The reason I ask is because the question gives plural 'equations
' so it implies more than one.

 

[CAF123]

I get that $A = M^{-1}K$. I now have to diagonalise A. In my previous problems I have simply put the ansatz of $\underline{x} = \underline{P}(a \cos (wt) + b \sin (wt))$ and this has allowed me to obtain the form $G\underline{P} = \lambda \underline{P}$.

Is it what I should do here? It seems appropraite given the form of the energy and eqn of motion.

 

[voko]

Why do you need to diagonalize it? Are you supposed to find a closed-form solution?

 

[CAF123]

Why do you need to diagonalize it? Are you supposed to find a closed-form solution?

Yes, I am supposed to find a general solution to the equation of motion. This is probably a bit late to ask but what is the underlying physics behind diagonalising a matrix?

 

[voko]

The physics is mostly in eigenvalues and eigenvectors. These correspond to "pure tones", so to speak, which any complex motion of the system can be made of.

Diagonalization is a mathematical device that makes use of these physical elements to render the problem in a form easier to deal with. One could also say that it transforms the problem from the "ad hoc" coordinates that we used to describe the problem initially, into "intrinsic" or "physical" coordinates.

 

[CAF123]

The physics is mostly in eigenvalues and eigenvectors. These correspond to "pure tones", so to speak, which any complex motion of the system can be made of.

Diagonalization is a mathematical device that makes use of these physical elements to render the problem in a form easier to deal with. One could also say that it transforms the problem from the "ad hoc" coordinates that we used to describe the problem initially, into "intrinsic" or "physical" coordinates.

Thanks voko, I can take the rest of the question from here. Now to try the next problem - a double pendulum... :)